4 Countability axioms

Transcription

1 4 COUNTABILITY AXIOMS 4 Countability axioms Definition 4.1. Let X be a topological space X is said to be first countable if for any x X, there is a countable basis for the neighborhoods of x. X is said to be second countable if there is a countable basis for the topology on X. Of course, second countable first countable. First countability is a mild assumption, while having a countable basis gives much more information about a space. A large class of first countable spaces are metric spaces: just take the balls at x with radius 1/n, n N (or rational). We proved the following proposition for metric topologies: Proposition 4.. Let X be first countable, A X, Y a topological space. Then (i) x A if, and only if, there is a sequence x n A, n N, such that x n x. (ii) f : X Y is continuous if, and only if, for any x n x, x n, x X, then f (x n ) f (x). Note that if we are given a countable basis at x, then there is always a decreasing countable basis: just take B 1 = B 1, B = B 1 B,..., B n = n i=1 B n - show it is basis. In general, a limit point is not necessarily the limit of some sequence, as the following examples show. Examples R J with J uncountable, with the product topology, take A to be the subset of all sequences x = (x α ) with x α = 1 for all but finite values of α, then the zero sequence is A but there is no sequence in A converging to 0 (since J is uncountable, for any sequence x n = (x nα ) A there must α J such that x nα = 1 for all n N). Hence R J with J uncountable is not first countable in the product topology, and also not metrizable.. R N the space of real sequence with the box topology and take A = {x = (x n ) : x n > 0}, show that the zero sequence is in A but there is no sequence in A converging to 0. Hence R N the space of real sequences with the box topology is not first countable, and also not metrizable. More examples: Examples A metric space (X, d): always first countable, not necessarily second countable (see examples below). Compact metric spaces are always second countable: take a finite subcover of {B(x, 1/n) : x X, n N} - see it is a basis. in fact, it suffices to be able to take a countable subcover of any open cover. 1

2 4 COUNTABILITY AXIOMS. Discrete space X: always first countable: {x} is a basis at x, and metrizable, second countable if, and only if, X is countable.. R n is second countable (hence first countable): ]p 1, q 1 [... ]p n, q n [, p i, q i Q, i = 1,..., n is a countable basis for the standard topology. 4. R N the space of real sequences, with the product topology, is second countable: the collection of sets of the form is a countable basis. ΠU n, U ni =]p i, q i [, p i, q i Q, i = 1,..., k, U n = R, n n i. 5. An uncountable set with the co-finite / co-countable topology: is not first countable: if B is a basis, x X, have {x} = B B B (it is T1 - see next section), if B is countable then X \ {x} is countable union of finite/countable sets, hence countable. 6. R with the topology given by slotted discs: a basic neighborhhod of x R is given by {x} B with B an open ball with straight lines through x removed: not first countable. 7. R J with J uncountable in the product topology, and also R N the space of real sequences with the box topology, are not first countable, and hence not metrizable. 8. R N ρ the space of real sequences, with the uniform topology: first countable, as it is a metric space. not second countable: {0, 1} N is an uncountable discrete subset, as ρ(x, y) = 1 for any sequences x y, hence B ρ (x, 1) = {x} is open, so any basis must be uncountable. Proposition 4.5. (i) subspaces of first / second countable spaces are first / second countable; (ii) countable products of first / second countable spaces are first / second countable (in product topology). (iii) open, continuous images of first / second countable spaces are first / second countable. Proof. Let B = {B n } n N be a countable basis (at x A or for the whole topology of X). Then {B A : B B} is also a basis. Let X = Π n N X n and B n be a basis for X n (at x n X n or for the whole topology of X n ). Then the collection of open sets of the form ΠU n such that U ni B ni for i = 1,..., p and U n = X n, n n i forms a basis. Let f : X Y be a continuous, open map. Then if B is a basis, also { f (B) : B B} is a basis. 14

3 4 COUNTABILITY AXIOMS Moreover, one can prove that a product of first / second countable spaces is first / second countable if, and only if, it is a countable product. It follows from (iii) that countability axioms are topological properties, ie, preserved by homeomorphisms (it is however not true in general that the continuous image of a first / second countable space is first / second countable). It is easy to check that in a space with a countable basis: (i) discrete subsets must be countable; (ii) collections of disjoint open sets must be countable. Recall that a subspace D X is dense if D = X. Proposition 4.6. If X has a countable basis, then (1) X has a countable dense subset, () any open cover of X has a countable subcover. Proof. Let B = {B n } n N be a countable basis. To prove (1), pick x n B n and let D = {x n : n N}. Then for any x X and any neighborhood U of x, there is B n U, hence D U and x D, that is D is dense. 4 To prove (), let U = {U α } be an open cover of X and for each n N, let U n = {U U : B n U}. If U n, pick U n U n and let U = {U n } n N U. Then for x X, exists U U and B n B such that x B n U, so U n is defined and x B n U n, and we conclude that U covers X. A space X that satisfies (1) is said to be separable and a space X that satisfies () is said to be Lindeloff. These conditions are usually weaker than being second countable, for instance R l is separable and Lindeloff but not second countable. However, if we a have a metric: Proposition 4.7. If X is metrizable, then second countable separable Lindeloff. Proof. If X is separable, and D is a dense countable set, then B = {B(a, 1/n) : a D} is countable and is a basis for the topology of X: for any open U, x U, there is n N such that B(x, 1/n) U and there is a B(x, 1/(n)), as D is dense. Then B(a, 1/(n)) U. If X is Lindeloff, take a countable subcover of the collection B(x, 1/n), show it is a basis. If X compact, it is Lindeloff. If it is compact and metrizable, then it is second countable (can see directly: since for any n N it can be covered by a finite number of ball of radius 1/n. The family of all those balls is a countable basis). Proposition 4.8. are Lindeloff. (i) open subspaces of separable spaces are separable; closed subspaces of Lindeloff A subset A X is discrete if the induced subspace topology is discrete, ie, {x} is open in A, x A. 4 This provides a way of constructing a dense subset from any basis: just pick one element from each basic set. 15

4 4 COUNTABILITY AXIOMS (ii) countable products of separable spaces are separable (in product topology). (iii) continuous images of separable/ Lindeloff spaces are separable/ Lindeloff. The next examples show that if X is not second countable, and not metrizable, one can get any combination of this separation properties: Examples R l. R l R l : first countable: [x, x + q[, q Q is a basis at x R not second countable: if B is a basis, let B x B such that B x [x, x + 1[. Then x y B x B y, as x = min B x min B y, hence B is uncountable. separable: Q is dense; Lindeloff (see Munkres). first countable, as R l is and separable, as R l is, not second countable, as R l {0} is a subspace, not Lindeloff: Let L = {(x, x) : x R} then L is closed and L [x, b[ [ x, c[= {(x, x)} (ie, the subspace topology in L is discrete; since L uncountable, R not second countable). l Cover R l by R \R and [a, b[ [ a, c[, a, b, c R, then there is no countable subcover, l hence not Lindeloff. Also shows: L closed, not separable, and L R separable. l. The ordered square I = [0, 1] [0, 1] with the dictionary order: 0 first countable not second countable: the collection of disjoint open sets {x} ]0, 1[, x ]0, 1[ is uncountable (or {x} {1/} is a discrete uncountable set). not separable it is Lindeloff - compact; Also shows [0, 1] ]0, 1[ not Lindeloff (open), since the collection {x} ]0, 1[, x [0, 1], has no countable subcover. 16

5 5 SEPARATION AXIOMS 5 Separation axioms Definition 5.1. Let X be a topological space, x, y X. X is T1 : if for x, y X, x y there exist neighborhoods U of x and V of y such that y U and x V. X is Hausdorff (or T) if for x, y X, x y there exist neighborhoods U of x and V of y such that U V =. X is regular if X is T1 5 and for x X, B X closed with x B, there exist open sets U, V such that x U, B V and U V =. X is normal if X is T1 and for A, B X closed with A B =, there exist open sets U, V such that A U, B V and U V =. Always have: normal regular Hausdorff T1 but the reverse implications NOT true in general (Hausdorff, not regular: R K ; T1, not Hausdorff: R K /K; regular, not normal: R l, or RJ, J uncountable) Remark In T1 spaces: x {y}, all x y. X is T1 {x} is closed, x X {x} = x U open U = B Bx B, B x basis of nbhds of x. In T1 spaces, x is limit point of A iff any nbhd of x intersects A in infinitely many points. (If F = (A \ {x}) U is finite, hence closed, then X \ F) is open nbhd of x that does not intersect A.). In Hausdorff spaces, there is unicity of limits of sequences and of extensions of continuous functions defined on dense subsets. Moreover: X is Hausdorff Λ = {(x, x) : x X} is closed in X X (note that U V = iff (U V) Λ =, so X is Hausdorff iff (x, y) Λ for x y, that is, Λ Λ). Examples X infinite with the cofinite topology is T1 but not Hausdorff. 6. R K Hausdorff, not regular: K is closed cannot be separated from 0.. R l is normal, R l not normal (but regular). 5 In R with topology given by ], a[, a R (is T0, not T1, not Hausdorff) any two closed sets intersect 6 It is the smallest T1 topology on X - check. 17

6 5 SEPARATION AXIOMS 4. if J uncountable R J not normal (hard to prove: see Ex..9 [Munkres] - see that N J not normal) The following result gives equivalent definitions of regular and normal spaces, that we will use quite often: Proposition 5.4. Let X be a T1 space. (i) X is regular for all x X and U open, x U, there exists open W, x W, such that W U. (ii) X is normal for any A X closed and U A open, there exists open W A such that W U. Proof. (i) U open, x U B = X \ U is closed, x B, hence X is regular iff for all x X and U open, x U, there exists open W, x W, and open V X \ U s.t. W V =. Then W V =, so W U. Conversely, if W U, then the open V = X \ W X \ U = B and U V =. (ii) is proved in a similar way. Proposition ) Products: if α J, X α is T1 / Hausdorff / regular, then X = Π α J X α is T1 / Hausdorff / regular. ) Subspaces: if X is T1 / Hausdorff / regular and A X, then A is T1 / Hausdorff / regular. For normal spaces, not true: R not normal, R l l normal so 1) fails and [0, 1] J normal, ]0, 1[ J not normal, so ) fails. However, if A X is closed and X is normal, then A is normal. In fact, the image of a normal space under a closed continuous map is normal. It is not true in general that images under continuous functions of Hausdorff / regular spaces are Hausdorff / regular, not even open or closed images. However, it is easy to see that any of the separation axioms are topological, in that they are preserved by homeomorphisms. Even if regular is in general weaker that normal, if we add a countability axiom then the two notions coincide. Theorem 5.6. Let X have a countable basis (ie, second countable). Then X regular X normal. Proof. Need to show that X regular and second countable then X is normal. Let B be a countable basis for X and A, B X be closed. For each x A, by regularity, can take a neighborhood U of x such that U B =. Since any such U contains a basis element and B is countable, we obtain a countable covering of X by sets U n such that U n B =, n N. We can do the same for the set B to obtain a covering of B by open sets V n such that V n A =, n N. Let U = n N U n, V = n N V n. Then U, V open, and U A, V B, but U and V need not be disjoint. Now let U n = U n \ n k=1 V k, V n = V n \ n k=1 U k and U = U n, V = V n. Then U, V open, U A, V B and u V =. 18

7 6 URYSOHN S LEMMA AND APPLICATIONS Note that we proved in fact that if X is Lindeloff and regular, then it is normal. Theorem 5.7. Let X be compact Hausdorff. Then X is normal. Proof. We have seen that in a Hausdorff space compact (disjoint) sets are separated by open sets. Since any closed subset of X is compact, as X is compact, normality follows. Corollary 5.8. Any locally compact Hausdorff space is regular. Proof. It is a subspace of a normal, hence regular, space hence it is regular. (Alternatively, any locally compact Hausdorff has a basis of neighborhoods with compact closure: given x X and U a neighborhood of x, there exists open B, with x B and B U, ie., X is regular.) A locally compact Hausdorff space needs not be normal: e.g., take ]0, 1[ J [0, 1] J with J uncountable. Then ]0, 1[ J is locally compact Hausdorff as it is open in [0, 1] J, which is compact Hausdorff (hence normal). But ]0, 1[ J R J not normal. Theorem 5.9. Let X be metrizable. Then X is normal. Proof. Let d be a metric inducing the topology on X, and A, B X be closed, A B =. Then for each a A and b B, since A B = and B A =, we can find ɛ a > 0 and ɛ b > 0 such that B(a, ɛ a ) B = and B(b, ɛ b ) A =. Let U = a A B(a, ɛ a /), V = b B B(b, ɛ b /). Then U, V open, A U, B V and U V = : for a A, b B, d(a, b) max(ɛ a, ɛ b ), but if there were z B(a, ɛ a /) B(b, ɛ b /) then d(a, b) d(a, z) + d(z, b) < ɛ a + ɛ b < max(ɛ a, ɛ b ). Examples R n is normal.. Any discrete space is normal - metrizable.. R N is normal - metrizable (any countable product of metrizable spaces is metrizable). 4. R J with the uniform topology is normal (not with product!) 6 Urysohn s Lemma and applications 6.1 Urysohn s Lemma One of the most important features of normal spaces is that normality is the suitable condition to prove the very useful Urysohn s Lemma. Theorem 6.1 (Urysohn s Lemma). Let X be a normal space, A, B X closed and disjoint. Then there exists continuous f : X [0, 1] s.t. f (A) = 0, f (B) = 1. 19

8 6.1 Urysohn s Lemma 6 URYSOHN S LEMMA AND APPLICATIONS (Of course, [0, 1] can be replaced by any interval [a, b].) Before we get to the proof, note that, for a T1 space, the existence of such a function f clearly yields normality: U = f 1 ([0, ɛ[ and V = f 1 (]1 ɛ, 1] are open disjoint (if ɛ 1/) sets that contain A and B, respectively. To prove this theorem, we have to use the the topology on X to construct from scratch such a continuous function f. The main idea is to construct a sequence of ordered neighborhoods of A that will be the level sets of f. We will use the alternative characterization of normal spaces: in a normal space, given A closed and U A open, there exists an open V A such that V U. Proof. We outline the proof (see the details in [Munkres]). Step 1) First we construct a sequence U q, for q [0, 1] Q such that U q is open and p < q U p U q. We do this as follows: let U 1 = X \ B and by normality, take U 0 open such that A U 0 and U 0 U 1. Let P = {q n : n N, q n [0, 1] Q}, for instance P = {1, 0, 1, 1,, 1 4, 4, 1 5, 5,...}. Then since U 0 U 1, U 0 closed, U 1 open, we can take U 1 open such that U 0 U 1 and U 1 U 1. We then proceed to define U 1 in between U 0 and U 1, then U in between U 1 and U 1, etc. U 0 open such that A U 0 and U 0 U 1 ; U 1 open such that U 0 U 1 and U 1 U 1 ; U 1 open such that U 0 U 1 and U 1 U 1 ; U open such that U 1 U 1 and U U 1 ;... We can show by induction that if we have a family U q defined for the first n elements of P then also U qn+1 is well-defined: just take its immediate predecessor p and sucessor q in P n = {q 0, q 1,..., q n }. Since p < q, U p U q ; let U qn+1 such that U p U qn+1 and U qn+1 U q. Then U p U q for all p < q, for all p, q P n+1, that is, for all p, q Q [0, 1]. For convenience, we define U q = for q Q, q < 0, and U q = X, for q Q, q > 1. Then U p U q for all p < q, p, q Q. Step ) For any x X, the set {p Q : x U p } is non-empty and bounded from below, hence we can define f (x) = inf{p Q : x U p }. Then f (A) = 0, and f (B) = 1. Step ) Show that f is continuous. Note that (1) x U r f (x) r: since if x U r then x U s, for s > r, hence f (x) r; 0

9 6 URYSOHN S LEMMA AND APPLICATIONS 6. Completely regular spaces () x U r f (x) r: since for s < r, U s U r, hence x U s, s < r and f (x) r. Let x 0 X such that f (x 0 ) ] a, b [. Take p, q Q such that x 0 ] p, q [ ] a, b [. Then, p < f (x 0 ) < q yields that x 0 U q \ U p (by (), f (x) < q x U q and by (1), f (x) > p x U p ). Moreover, x U q f (x) q, x U p f (x) p. It follows that for any x U = U q \ U p open, f (x) [p, q] hence f (U) ] a, b [ and f is continuous. Corollary 6.. Let X be a normal space, A closed and U A open. Then there exists continuous f : X [0, 1] s.t. f (A) = 1, f (x) = 0, x U. 7 Proof. Just take B = X \ U in Urysohn s lemma. It follows from Urysohn s Lemma that X is normal X is T1 and for A, B X closed, there is f : X [0, 1] continuous such that f (A) = 0 and f (B) = Completely regular spaces Definition 6.. X is completely regular if X is T1 and for x X, B X closed with x B, there if f : X [0, 1] continuous such that f (x) = 0, f (B) = 1. A completely regular space is regular: given x X and A closed with x A, let U = f 1 ([0, 1/[ and V = f 1 (]1/, 1]). We always have normal completely regular regular but the reverse implications are also NOT true in general: R is completely regular (being l the product of completely regular spaces) but not normal. Examples of regular spaces that are not completely regular are more sophisticated (see Tychonoff cork screw example, or [Munkres] Ex..11). This new axiom is better behaved than normality: Proposition ) Products: if α J, X α is completely regular, then X = Π α J X α is completely regular. ) Subspaces: if X is completely regular and A X, then A is completely regular. It follows that any subspace of a normal space is completely regular. In particular, any subspace of a compact, Hausdorff space is completely regular, hence any locally compact Hausdorff space is completely regular. We see now that completely regular spaces can always be embedded in a compact Hausdorff space, which leads to a characterization of spaces that have a compactification, and is a step in proving Urysohn s metrization theorem. First note that we can give equivalent definition of a completely regular space using neighborhoods, instead of closed sets: X is completely regular if, and only if, X is T1 and given x 0 U open, there exists a continuous function f : X [0, 1] such that f (x 0 ) = 1 and f (x) = 0, x U. 7 f is said to have support in U. 1

10 6. Completely regular spaces 6 URYSOHN S LEMMA AND APPLICATIONS Theorem 6.5 (Embedding theorem). Let X be T1. If there is a family of continuous functions f α : X [0, 1] such that for x 0 U open, there exists α such that f α (x 0 ) > 0 and f α (X \ U) = 0, then F : X [0, 1] J, F(x) = ( f α (x)) α J is an embedding. Proof. First, F is continuous: in the product topology F is continuous iff π α F = f α is continuous for all α J. Then F is injective: if x y, take a neighborhood U of x such that y U (X is T1). Then there is α J such that f α (x) > 0 and f α (y) = 0, hence (F(x)) α (F(y)) α F(x) F(y). Now see that F is open: let U X be open, x 0 U and z 0 = F(x 0 ) F(U). We show that there is a neighborhhod of z 0 contained in F(U), hence F(U) is open. Take α such that f α (x 0 ) > 0, ie, (z 0 ) α > 0, and f α (X \ U) = 0. Let W = π 1 α (]0, 1]) F(X), open in F(X). Then z 0 W and z W π α (z) > 0 z F(X) z = F(x) π α (F(x)) = f α (x) > 0 x U. Hence, W F(U). Since the conditions of the embedding theorem are equivalent to X being completely regular, and by Tychonoff s theorem, [0, 1] J is compact, it follows: Corollary 6.6. X is completely regular if, and only if, X is homeomorphic to a subspace of a compact, Hausdorff space. In particular, any metrizable (hence normal) space embeds in a compact, Hausdorff space. Remark 6.7. We saw in section that a compactification of a space X is a compact Hausdorff space Y such that X Y and X = Y. There can be many compactifications to a given (necessarily completely regular) space. We saw that if X is locally compact, then there is a minimal compactification, adding one point at infinity. Now any embedding h : X Z with Z compact, Hausdorff, defines a compactification of X: take X 0 = h(x) X and Y 0 = h(x), which is compact, Hausdorff (being closed). Let A be a set such that there is a bijection k : A Y 0 \ X 0 and define Y = X A. Then define H : Y Y 0 such that H(x) = h(x), x X, h(x) = k(x), x A. If we endow Y with the topology given by U open iff H(U) open, we get that H is a homeomorphism, Y is compact Hausdorff and X Y is a subspace. Moreover, H(X) = H(X) = h(x) = Y 0, hence X = Y. (Note that H : Y Z is an embedding that extends h.) The results above show that X is completely regular iff it has a compactification.

11 6 URYSOHN S LEMMA AND APPLICATIONS 6. Urysohn s Metrization theorem 6. Urysohn s Metrization theorem We will now see Urysohn s metrization theorem. Recall that a countable product of metric spaces is always metrizable: if X = Π n N X n and d n (a) = min{d n (a), 1} is the standard bounded metric associated to the metric d n on X n, then we can define a metric (inducing the topology) on X for instance by D(x, y) := sup n N d n (x n, y n ). n Theorem 6.8 (Urysohn s metrization theorem). Let X be regular and have countable basis. Then X is metrizable. (Note that metrizable regular, but not necessarily second countable. eg discrete uncountable space, or the space of sequence R N in the uniform topology.) Proof. We will construct an embedding of X in [0, 1] N with the product topology, which yields that X, being a subspace of a metrizable space, is also metrizable. We showed in Theorem 5.6 that a regular, second countable space is normal. We will use Urysohn s lemma, and the countable basis, to construct a countable family satisfying the conditions of the Embedding theorem. Let B = {B n } n N be a countable basis for the topology on X. Let x 0 X, U a neighborhood of x 0 and B n B such that x 0 B n U. By regularity, we can take B m such that B m B n U. Now apply Urysohn s lemma (in fact, Corollary 6.) to B m B n to get that there is a continuous function g n,m : X [0, 1], g n,m (B m ) = 1, g n,m (X \ B n ) = 0, in particular, g n,m (x 0 ) = 1 and g n,m (x) = 0, x U. We obtain in this way a countable family of function satisfying the conditions of the Embedding theorem. Hence X can be embedded in [0, 1] N, and therefore it is metrizable. Corollary 6.9. Let X be second countable. Then X metrizable X is regular X is normal. Proof. We have proved already that for a second countable space, regular normal, and that if X metrizable then normal. The remaining follows form the Metrization theorem. Corollary X regular and second countable X metrizable and separable Proof. Recall that if X is metrizable, then second countable separable. Corollary Let X be compact Hausdorff. Then X metrizable second countable. Example 6.1. Recall that a n-manifold M is a Hausdorff, second countable space where each point has a neighborhood diffeomorphic to an open subset of R n. Then, since the closed ball in R n is compact (Heine-Borel s theorem), we can show that M is locally compact, hence regular. It follows that any n-manifold is metrizable. Another nice application of Urysohn s Lemma is a short proof that any compact n- manifold can be embedded in R N, for some N (see section 6 - [Munkres]).

12 6.4 Tietze extension theorem 6 URYSOHN S LEMMA AND APPLICATIONS 6.4 Tietze extension theorem Our final application of Urysohn s lemma is an extension theorem. Extending continuous functions is a useful tool in many applications of topology. A consequence of Urysohn s lemma is the following important theorem, that says that in normal spaces, real functions on a closed subset can be always extended to the whole space. (It is clear that on open sets that is not always possible, e.g, f (x) = 1/x on R +.) Theorem 6.1 (Tietze extension theorem). Let X be a normal space, A X be closed. Then (i) f : A [a, b] continuous can be extended to g : X [a, b] continuous; (ii) f : A R continuous can be extended to g : X R continuous. In fact, this theorem is equivalent to Urysohn s lemma, in that, in one direction, we use it in the proof, and conversely, if (i) or (ii) holds then f (A) := 0, f (B) := 1 for any disjoint, closed A, B X can be extended to X, since it is continuous (Pasting lemma for instance). In particular, assuming X is T1, it follows from Tietze s theorem that it is normal Proof. The main idea is to construct a sequence s n of continuous functions, which converges uniformly to some function g with g = f on A. Uniform convergence guarantees that g is continuous. Let us consider the case (i), can assume that [a, b] = [ 1, 1]. (Then (ii) will follow, identifying R ] 1, 1[ and using (i).) Let B 1 = f 1 ([ 1, 1 ]) and C 1 = f ([ 1 1, 1]). Then B 1, C 1 are closed in A, hence in X, as A is closed, and disjoint. By Urysohn s lemma, there exists [ g 1 : X 1, 1 ], g 1 (B 1 ) = 1, g 1(C 1 ) = 1. In particular: (i) g 1 (x) 1, for x X, (ii) f (a) g 1 (a), for a A: note that [ 1, 1] = [ 1, 1 ] [ 1, 1 ] [ 1, 1] and each of these sub-intervals has length /, so it suffices to show that f (a) and g 1 (a) are in the same sub-interval. Now have cases: a B 1 : then f (a) [ 1, 1 ] and g 1(a) = 1 ; a C 1 : then f (a) [ 1, 1] and g 1 (a) = 1 ; a A \ (B 1 C 1 ): then f (a), g 1 (a) [ 1, 1 ]. Let s 1 = g 1. Now apply the step above to the function f g 1 : A [, ] to obtain a function g : X [ 9, 9] such that Let s = g 1 + g. ( ) f (a) g 1 (a) g (a), a A. 4

13 6 URYSOHN S LEMMA AND APPLICATIONS 6.4 Tietze extension theorem We can proceed by induction to define functions g n : X writing s n = n i=1 g i, we get ( ) n f (a) s n (a), a A. [ ( ) 1 n 1 (, 1 n 1 ] ) such that, In particular, for a A, s n (a) f (a), that is, s n converges pointwise to f in A. ( ) Moreover, since g n (x) 1 n 1, by the comparison test for infinite series, for each x X, s n (x) = n i=1 g i(x) converges and we define g(x) = n=1 g n (x) = lim n s n (x). We have that g = f on A. Need only show that g is continuous, which will follow once we show that the convergence is uniform (exercise). In fact, the fact that the sequence s n converges uniformly (can be proved directly but) follows from the fact that it is a Cauchy sequence and that R is complete. We will see these notions next. 5